# NCERT Solutions for Class 11 Chemistry Chapter 8: Redox Reactions

## NCERT Solutions for Class 11 Chemistry Chapter 8 Free PDF Download

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8.1. Assign oxidation number to the underlined elements in each of the following species:

(a) NaH2PO4

(c) H4P2O7

(e) CaO2

(g) H2S2O7

(b) NaHSO4

(d) K2MnO4

(f) NaBH4

(h) KAI(SO4)2.12H2O

Ans. (a) NaH2PO4

Let the oxidation number of P be x.

We know that,

Oxidation number of Na = +1

Oxidation number of H = +1

Oxidation number of O = –2

$$\Rarr\space\text{Na}^{\normalsize+1}\text{H}_2^{\normalsize+1}\text{P}^{x}\text{O}_4^{\normalsize-1}$$

Then, we have

1(+1) + 2(+1) +1(x) + 4(–2) = 0

⇒ 1 + 2 + x – 8 = 0

⇒ x = +5

Hence, the oxidation number of P is +5.

(b) NaHSO4

Let the oxidation number of S be x

We know that:

Oxidation number of Na = +1

Oxidation number of H = +1

Oxidation number of O = –2

$$\text{Na}^{\normalsize+1}\text{H}^{\normalsize+1}\text{S}^{x}\text{O}_4^{\normalsize-2}$$

Then, we have

1(+1) + 1(+1) +1(x) + 4(–2) = 0

⇒ 1 + 1 + x – 8 = 0

⇒ x = +6

Hence, the oxidation number of S is +6.

(c) H4P2O7

Let the oxidation number of P be x

We know that

Oxidation number of H = +1

Oxidation number of O = –2

$$\text{H}_4^{\normalsize+1}\text{P}_2^{x}\text{O}_7^{\normalsize-2}$$

Then, we have

4(+1) + 2(x) + 7(–2) = 0

⇒ 4 + 2x – 14 = 0

⇒ 2x = +10

⇒ x = +5

Hence, the oxidation number of P is +5.

(d) K2MnO4

Let the oxidation number of Mn be x

We know that

Oxidation number of K = +1

Oxidation number of O = –2

$$\text{K}_2^{\normalsize+1}\text{Mn}^{x} \text{O}^{\normalsize-2}_4$$

Then, we have

2(+1) +x + 4(–2) = 0

⇒ 2 + x – 8 = 0

⇒ x = +6

Hence, the oxidation number of Mn is +6.

(e) CaO2

Let the oxidation number of O be x

We know that

Oxidation number of Ca = +2

$$\text{Ca}^{\normalsize+2}\text{O}_2^{x}$$

Then, we have

(+2) + 2(x) = 0

⇒ 2 + 2x = 0

⇒ x = –1

Hence, the oxidation number of O is –1.

(f) NaBH4

Let the oxidation number of B be x

We know that

Oxidation number of Na = +1

In NaBH4, H is present as hydride ion. Therefore, its oxidation number is –1.

$$\text{Na}^{\normalsize+1}\text{B}^{x}\text{H}_4^{\normalsize-1}$$

Then, we have

1(+1) + 1(x) + 4(–1) = 0

⇒ 1 + x – 4 = 0

⇒ x = +3

Hence, the oxidation number of B is +3.

(g) H2S2O7

Let the oxidation number of S be x

We know that

Oxidation number of H = +1

Oxidation number of O = –2

$$\text{H}_{2}^{\normalsize+1}\text{S}_{2}^{x}\text{O}_{7}^{\normalsize-2}$$

Then, we have

2(+1) + 2(x) + 7(–2) = 0

⇒ 2 + 2x – 14 = 0

⇒ 2x = 12

⇒ x = +6

Hence, the oxidation number of S is +6.

(h) KAI(SO4)2.12H2O

Let the oxidation number of S be x

We know that

Oxidation number of K = +1

Oxidation number of Al = +3

Oxidation number of O = –2

Oxidation number of H = +1

$$\text{K}^{\normalsize+1}\text{AI}^{3+}.\text{(SO}_4)_2^{x-2}.12\text{H}_2^{\normalsize+1}\text{O}^{\normalsize-2}$$

Then, we have

1(+1) + 1(+3) + 2(x) + 8(–2) + 24(+1) + 12(–2) = 0

⇒ 1 + 3 + 2x – 16 + 24 – 24 = 0

⇒ 2x = 12

⇒ x = +6

Hence, the oxidation number of S is +6.

8.2. What are the oxidation number of the underlined elements in each of the following and how do you rationalise your results?

(a) KI3

(c) Fe3O4

(e) CH3COOH

(b) H2S4O6

(d) CH3CH2OH

Ans. (a) KI3

Let x be the oxidation number of I

Oxidation number of K = +1

Then

1(+1) + 3(x) = 0

1 + 3x = 0

$$x=-\frac{1}{3}\\\text{Hence, the average oxidation number of I is}\space -\frac{1}{3}.$$

However, oxidation number cannot be fractional. Therefore, we will have to consider the structure of KI3 to find the oxidation states.

In a KI3 molecule, an atom of iodine form a coordinate covalent bond with an iodine molecule.

$$\text{K}^{\normalsize+^{\normalsize+1}}[\text{I}^{0}\text{\normalsize-I}^{0} \larr \text{I}^{\normalsize-1}]^{\normalsize-}$$

Hence, in a KI3 molecule, the oxidation number of the two I atoms forming the I2 molecule is 0, whereas the oxidation number of the I atom forming the coordinate bond is –1.

(b) H2S4O6

Let x the oxidation number of S.

Oxidation number of H = +1

Oxidation number of O = –2

$$\text{H}_{2}^{\normalsize+1}\text{S}_{4}^{x}\text{O}_{6}^{\normalsize-2}$$

Now, 2(+1) + 4(x) + 6(–2) = 0

⇒ 2 + 4x – 12 = 0

⇒ 4x = 10

$$\Rarr\space x=+2\frac{1}{2}$$

However, oxidation number cannot be fractional. Hence, S must be present in different oxidation sates in the molecule.

(c) Fe3O4

Let x be the oxidation number of Fe oxidation number of O = –2

Then, 3(x) + 4(–2) = 0

$$3x – 8 = 0, x =\frac{8}{3}\\\text{On taking the oxidation number of O as –2, the oxidation number of Fe is found to be}+2\frac{2}{3}.$$

However, oxidation number cannot be fractional.

Here, one of the three Fe atoms exhibits the oxidation number of +2 and the other two Fe atoms exhibit the oxidation number of +3.

$$\text{Fe}_{3}^{\normalsize+2}\text{O},\text{Fe}_2^{\normalsize+3}\text{O}_3$$

Hence, the oxidation number of Fe in Fe3O4 is +2 and 3.

(d) CH3CH2OH

Let x be the oxidation number of C

Oxidation number of O = –2

Oxidation number of H = +1

Then,

$$\text{C}_{2}^{x}\text{H}_{6}^{\normalsize+1}\text{O}^{\normalsize-2}$$

2(x) + 6(+1) + 1(–2) = 0

⇒ 2x + 6 –2 = 0

⇒ x = –2

Hence, the oxidation number of C is –2.

(e) CH3COOH

Let x be the oxidation number of C

Oxidation number of O = –2

Oxidation number of H = +1

Then,

$$\text{C}_{2}^{x}\text{H}_{4}^{\normalsize+1}\text{O}_{2}^{\normalsize-2}$$

2(x) + 4(+1) + 2(–2) = 0

⇒ 2x + 4 – 4 = 0

⇒ x = 0

However, 0 is average oxidation number of C. The two carbon atoms present in this molecule are present in different environments. Hence, they cannot have the same oxidation number.

Thus, C exhibits the oxidation states of +2 and –2 in CH3COOH.

8.3. Justify that the following reactions are redox reactions:

(a) CuO(s) + H2(g) → Cu(s) + H2O(g)

(b) Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2/(g)

(c) 4BCl3(g) + 3LiAlH4(s) → 2B2H6(g) + 3LiCl(s) + 3AlCl3(s)

(d) 2K(s) + F2(g) → 2K+F(s)

(e) 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)

Ans. (a) For the given reaction,

$$\text{Cu}^{\normalsize+2}\text{O(s)}^{\normalsize-2}\text{+\text{H}}_{2}^{0}\text{(g)}\xrightarrow{}\text{Cu}(\text{s})^{0}+\text{H}_{2}^{\normalsize+1}\text{O}^{\normalsize-2}\text{g}$$

Oxidation number of Cu and O in CuO is +2 and –2 respectively.

Oxidation number of H2 is 0.

Oxidation number of Cu is 0.

Oxidation number of H and O in H2O is +1 and –2 respectively.

Further, the oxidation number of Cu decreases from +2 in CuO to 0 in Cu but that of H increases from 0 in H2 to +1 in H2O. Thus, CuO is reduced to Cu but H2 is oxidised to H2O. Thus, the given reaction is a redox reaction.

(b) For the given reaction,

$$\text{Fe}_{2}^{\normalsize+3}\text{O}_{3}^{\normalsize-2}\text{(s)}+3\text{C}^{+2}\text{O}^{\normalsize-2}(\text{g})\xrightarrow{}2\text{Fe}^{0}\text{(s)}+3\text{C}^{\normalsize+4}\text{O}_{2}^{\normalsize-2}\text{(g)}$$

Oxidation number of Fe and O in Fe2O3 is +3 and –2 respectively.

Oxidation number of C and O in CO is +2 and –2 respectively.

Oxidation number of Fe is 0.

Oxidation number of C and O in CO2 is +4 and –2 respectively.

Further, the oxidation number of Fe decreases from +3 in Fe2O3 to 0 in Fe which means Fe2O3 is reduced to Fe. Whereas, the oxidation number of C is increased from 0 to +2 in CO to +4 in CO2 which means CO is oxidised CO2. Thus, the given reaction is a redox reaction.

$$\text{(c)}\space4 \text{B}^{+3}\text{Cl}_{3}^{\normalsize-1}\text{(g)}+\text{Li}^{\normalsize+1}\text{Al}^{\normalsize+3}\text{H}_{4}^{\normalsize-1}\text{(s)}\xrightarrow{}2\text{B}_{2}^{\normalsize-3}\text{H}_{6}^{+1}\text{(g)}+3\text{Li}^{+1}\text{Cl}^{-1}\text{(s)}+3\text{Al}^{\normalsize+3}\text{Cl}^{\normalsize-1}_{3}\text{(s)}$$

Oxidation number of B and Cl in BCl3 is +3 and –1 respectively.

Oxidation number of Li, Al and H in LiAlH4 is +1, +3 and –1 respectively.

Oxidation number of B and H in B2H6 is –3 and +1 respectively.

Oxidation number of Li and Cl in LiCl is +1 and –1 respectively.

Oxidation number of Al and Cl in AlCl3 is +3 and –1 respectively.

Further, the oxidation number of B is decreased from +3 in BCl;3 to –3 in B2H6 which means BCl3 is reduced to B2H6. Whereas, the oxidation number of H is increased from –1 in LiAlH4 to +1 in B2H6 which means LiAlH4 is oxidised to B2H6. Thus, the given reaction is redox reaction.

(d) For the given reaction,

$$\text{2K(s)}^{0}+\text{F}_{2}^{0}\text{(g)}\xrightarrow{}2\text{K}^{\normalsize+1}\text{F}^{\normalsize-1}\text{(s)}$$

Oxidation number of K is 0.
Oxidation number of F is 0.
Oxidation number of K and F in KF is +1 and –1 respectively.
Further, the oxidation number of K is increased from 0 in K to +1 in KF which means K is oxidised to KF. Whereas, the oxidation number of F is decreased from 0 in F2 to –1 in KF which means F2 is reduced to KF. Thus, the given reaction is a redox reaction.

(e) For the given reaction,

$$4\text{N}^{\normalsize-3}\text{H}_{3}^{+1}\text{(g)}+5\text{O}_{2}^{0}\text{(g)}\xrightarrow{}4\text{N}^{\normalsize+2}\text{O}^{\normalsize-2}\text{(g)}+6\text{H}_{2}^{\normalsize+1}\text{O (g)}$$

Oxidation number of O2 is 0.

Oxidation number of N and O in NO is +2 and –2 respectively.

Oxidation number of H and O in H2O is +1 and –2 respectively.

Further, the oxidation number of N is increased from –3 in NH3 to +2 in NO. Whereas, the oxidation number of O2 is decreased from 0 in O2 to –2 in NO and H2O which means O2 is reduced. Thus, the given reaction is a redox reaction.

8.4. Fluorine reacts with ice and results in the change:

H2O(s) + F2(g) → HF(g) + HOF(g)

Justify that this reaction is a redox reaction.

Ans. Let us write the oxidation number of each atom involved in the given reaction above its symbol as:

$$\text{H}_{2}^{\normalsize+1}\text{O}^{\normalsize-2}+\text{F}_{2}^{0}\xrightarrow{}\text{H}^{\normalsize+1}\text{F}^{\normalsize-1}+\text{H}^{\normalsize+1}\text{O}^{\normalsize-2}\text{F}^{\normalsize+1}$$

In the above reaction,

Oxidation number of H and O in H2O is +1 and –2 respectively.

Oxidation number of F2 is 0.

Oxidation number of H and F in HF is +1 and –1 respectively.

Oxidation number of H, O and F in HOF is +1, –2 and +1 respectively.

Here, we have observed that the oxidation number of F increases from 0 in F2 to +1 in HOF. Also, the oxidation number decreases from 0 in F2 to –1 in HF. Thus, in the above reaction, F is both oxidised and reduced. Hence, the given reaction is a redox reaction.

8.5. Calculate the oxidation number of sulphur, chromium and nitrogen in H2SO5, Cr2O72– and NO3. Suggest structure of these compounds. Count for the fallacy.

Ans. (i) Let x be the oxidation number of S

Oxidation number of O = –2

Oxidation number of H = +1

$$\text{H}_{2}^{\normalsize+1}\text{S}^{x}\text{O}^{\normalsize-2}_{5}$$

2(+1) + 1(x) + 5(–2) = 0

⇒ 2 + x – 10 = 0

⇒ x = + 8

However, the oxidation number of S cannot be +8. S has six valence electrons. Therefore, the oxidation number of S cannot be more than +6.

The structure of H2SO5 is shown as follows:

Now, 2(+1) + 1(x) + 3(–2) + 2(–1) = 0

⇒ 2 + x – 6 – 2 = 0

⇒ x = +6

Therefore, the O. N. of S is +6.

(ii) Let x be the oxidation number of Cr.

Oxidation number of O = –2

$$\text{Cr}_{2}^{x}\text{O}_{7}^{2-^{2-}}$$

Therefore, each of the two Cr atoms exhibits the oxidation number of +6.

(iii) Let x be the oxidation number of N

Oxidation number of O = –2

$$\text{N}^{x}\text{O}_{3}^{\normalsize- ^{2\normalsize-}}$$

1(x) + 3(–2) = –1

⇒ x – 6 = –1

⇒ x = +5

Here, there is no fallacy about the oxidation number of N in NO3–.

The structure of NO3 is shown as follows:

Therefore N atom exhibits the oxidation number of +5.

8.6. Write formulas for the following compounds:

(a) Mercury(II) chloride

(b) Nickel(II) sulphate

(c) Tin(IV) oxide

(d) Thallium(I) sulphate

(e) Iron (III) sulphate

(f) Chromium(III) oxide

Ans. (a) Mercury (II) chloride : HgCl2

(b) Nickel (II) sulphate : NiSO4

(c) Tin (IV) oxide : SnO2

(d) Thallium (I) sulphate : Tl2SO4

(e) Iron (III) sulphate : Fe2(SO4)3

(f) Chromium (III) oxide : Cr2O3

8.7. Suggest a list of the substances where carbon can exhibit oxidation states from –4 to +4 and nitrogen from –3 to +5.

Ans. The substances where carbon can exhibit oxidation states from –4 to +4 and nitrogen can exhbit oxidation state from –3 to +5 are listed in the table below :

 Compound Oxidation number of Carbon Compound Oxidation number of Nitrogen CH4 -4 NH3 -3 CH3CH3 -3 NH2-NH2 -2 CH2=CH2or CH3Cl -2 NH=NH -1 CH=CH -1 N=N 0 CH2Cl2 or C6H12O6 0 N2O +1 C2Cl2 or C6Cl6 +1 NO +2 CO or CHCl3 +2 N2O3 +3 C2Cl6 or (COOH)2 +3 N2O4 +4 CO2 or CCl4 +3 N2O5 +5

8.8. While sulphur dioxide and hydrogen peroxide can act as oxidising as well as reducing agents in their reactions, ozone and nitric acid act only as oxidants. Why ?

Ans. In sulphur dioxide (SO2), the oxidation number (O.N.) of S is +4 and the range of the oxidation number that S can have is from +6 to –2.
Therefore, SO2 can act as an oxidising as well as a reducing agent.

In hydrogen peroxide (H2O2), the oxidation number of Ois –1 and the range of the oxidation number that O can have is from 0 to –2. O can sometimes also attain the oxidation numbers +1 and +2.

Hence, H2O2 can act as an oxidising as well as a reducing agent.

In ozone (O2), the oxidation number of O is zero and the ragne of the oxidation number that O can have is from 0 to –2. Therefore, the oxidation number of O can only decrease in this case. Hence, O3 acts only as an oxidant.

In nitric acid (HNO3), the oxidation number of N is +5 and the range of the oxidation number that N can have is from +5 to –3. Therefore, the oxidation number of N can only decrease in this case. Hence, HNO3 acts only as an oxidant.

8.9. Consider the reactions:

$$\text{(a)}\space 6\text{CO}_2\text{(g)} + 6\text{H}_2\text{O(l)} \xrightarrow{}\text{C}_6\text{H}_{12}\text{O}_6(\text{aq}) + 6\text{O}_2(\text{g})\\\text{(b) O}_3\text{(g) + H}_2\text{O}_2\text{(l)} \xrightarrow{} \text{H}_2\text{O(l)} + 2\text{O}_2\text{(g)}$$

Why it is more appropriate to write these reactions as :

$$\text{(a)}\space 6\text{CO}_2\text{(g)} + 12\text{H}_2\text{O(l)}\xrightarrow{}\text{C}_6\text{H}_{12}\text{O}_{6}\text{(aq)}+6\text{H}_{2}\text{O}(1)+6\text{O}_2\text{(g)}\\\text{(b) \space O}_{3}\text{(g)}+\text{H}_{2}\text{O}_{2}\xrightarrow{}\text{H}_{2}\text{O(1)}+\text{O}_2\text{(g)}+\text{O}_{2}\text{(g)}$$

Also suggest a technique to investigate the path of the above (a) and (b) redox reactions.

Ans. (a) The process of photosynthesis involves two steps.

Step 1: H2O decomposes to give H2 and O2.

Step 2: The H2 produced in step 1 reduces CO2, thereby producing glucose (C6H12O6) and H2O.

$$6\text{CO}_2\text{(g)} + 12\text{H}_2\text{(g)}\xrightarrow{} \text{C}_6\text{H}_{12}\text{O}_6\text{(s)} + 6\text{H}_2\text{O(l)}$$

Now, the net reaction of the process is given as:

$$2\text{H}_{2}\text{O}(1)\xrightarrow{}2\text{H}_{2}\text{(g)}+\text{O}_{2}\text{(g)}×6\\6\text{CO}_{2}\text{(g)}+12\text{H}_{2}\text{O\space(g)}\xrightarrow{}\text{C}_{6}\text{H}_{12}\text{O}_{6}\text{(g)}+6\text{H}_{2}\text{O\space(g)}\\6\text{CO}_{2}\text{(g)}+12\text{H}_{2}\text{O}(1)\xrightarrow{}\text{C}_{6}\text{H}_{12}\text{O}_{6}\text{(s)}+6\text{H}_{2}\text{O(1)}+6\text{O}_{2}\text{(s)}$$

It is more appropriate to write the reaction as given above because it emphasises that 12 water (H2O) molecules are used per molecule of carbohydrate formed and 6H2O are produced during the process of photosynthesis.

The path of this reaction can be investigated by using radioactive H2O218 in place of H2O.

(b) O2 is produced from each of the two reactants O3 and H2O2. For this reason, O2 is written twice.

The given reaction involves two steps. First, O3 decomposes to form O2 and O. In the second step, H2O2 reacts with the O produced in the first step, thereby producing H2O and O2.

$$\space\space\space\space\space\space\qquad\qquad\qquad\text{O}_{3}\text{(g)}\xrightarrow{}\text{O}_{2}\text{(g)}+\text{O(g)}\\\frac{\text{H}_{2}\text{O}_{2}(1)+\text{O(g)}\xrightarrow{}\text{H}_{2}\text{O}(1)+\text{O}_{2}\text{(g)}}{\\\text{H}_{2}\text{O}_{2}(1)+\text{O}_{3}\text{(g)}\xrightarrow{}\text{H}_2\text{O}(1)+\text{O}_2\text{(g)}+\text{O}_{2}\text{(g)}}$$

The path of this reaction can be investigated by using H2O218 or O318.

8.10. The compound AgF3 is unstable compound. However, if formed, the compound acts as a very strong oxidising agent. Why ?

Ans. The oxidation state of Ag in AgF2 is +2. But, +2 is an unstable oxidation state of Ag. Therefore, whenever AgF2 is formed, silver readily accepts an electron to form Ag+. This helps to bring the oxidation state of Ag down from +2 to a more stable of +1. As a result, AgF2 acts as a very strong oxidising agent.

8.11. Whenever a reaction between an oxidising agent and a reducing agent is carried out, a compound of lower oxidation state is formed if the reducing agent is in excess and a compound of higher oxidation state is formed if the oxidising agent is in excess. Justify this statement giving three illustrations.

Ans. Whenever a reaction betwen an oxidising agent and a reducing agent is carried out, a compound of lower oxidation state is formed if the reducing agent is in excess and a compound of higher oxidation state is formed if the oxidising agent is in excess. This can be illustrated as follows :
(i) P4 and F2 are reducing and oxidising agents respectively.

If an excess of P4 is treated with F2, then PF3 will be produced, wherein the oxidation number (O.N.) of P is +3.

$$\text{P}_{4}(\text{excess})+\text{F}_{2}\xrightarrow{}\text{P F}_{3}^{+3}$$

However, if P4 is treated with an excess of F2, then PF5 will be produced, wherein the oxidation number of P is +5.

$$\text{P}_4 + \text{F}_2(\text{excess})\xrightarrow{}\text{P F}_3^{\normalsize+5}$$

(ii) K acts as a reducing agent, whereas O2 is oxidising agent.

If an excess of K reacts with O2, then K2O will be formed, wherein the oxidation number of O is –2.

$$4\text{K (excess)}+\text{O}_{2}\xrightarrow{}2 \text{K}_2\text{O}^-2$$

However, if K reacts with an excess of O2, then K2O2 will be formed, where in the oxidation number of O is –1.

$$2\text{K}+\text{O}_{2}\text{(excess)}\xrightarrow{}\text{K}_{2}\text{O}_{2}^{\normalsize-1}$$

(iii) C is a reducing agent, while O2 acts as an oxidising agent.
If an excess of C is burnt in the presence of insufficient amount of O2, then CO will be produced, wherein the oxidation number of C is +2.

$$2\text{C(excess) + O}_2\xrightarrow{}2\text{CO}^{\normalsize+2}$$

On the other hand, if C is burnt in an excess of O2, then CO2 will be produced, where the oxidation number of C is +4.

$$\text{C} + \text{O}_2(\text{excess}) \xrightarrow{} \text{CO}_2^{\normalsize+4}$$

8.12. How do you count for the following observations?

(a) Through alkaline potassium permanganate and acidic potassium permanganate both are used as oxidants, yet in the manufacture of benzoic acid from toluene we use alcoholic potassium permanganate as an oxidant. Why ? Write a balanced redox equation for the reaction.

(b) When concentrated sulphuric acid is added to an inorganic mixture containing chloride, we get colourless pungent smelling gas HCl, but if the mixture contains bromide then we get red vapour of bromine. Why ?

Ans. (a) In the manufacture of benzoic acid from toluene, alcoholic potassium permanganate is used as an oxidant becaue of the following reasons.

(i) In a neutral medium, OHions are produced in the reaction itself. As a result, the cost of adding an acid or a base can be reduced.

(ii) KMnO4 arnd alcohol are homogeneous to each other since both are polar. Toluene and alcohol are also homogeneous to each other because both are organic compounds.

Reactions can proceed at a faster rate in a homogeneous medium than in a heterogeneous medium. Hence, in alcohol, KMNo4 and toluene can react at a faster rate.

The balanced redox equation for the reaction in a neutral medium is give as below :

(a) When conc. H2SO4 is added to an inorganic mixture containing bromide, initially HBr is produced. HBr, being a strong reducing agent reduces H2SO4 to SO2 with the evolution of red vapour of bromide.

$$2\text{NaBr} + 2\text{H}_2\text{SO}_4\xrightarrow{} 2\text{NaHSO}_4 + 2\text{HBr}\\2\text{HBr +H}_2\text{SO}_{4}\xrightarrow{}\underset{\text{(red vapour)}}{\text{Br}_{2}+\text{SO}_{2}+2\text{H}_{2}\text{O}}$$

But, when conc. H2SO4 is added to an inorganic mixture containing chloride, a pungent smelling gas (HCl) is evolved because a stronger acid displaces a weaker acid from its salt. HCl, being a weak reducing agent, cannot reduce H2SO4 to SO2.

$$\underset{(\text{Stronger acid})}{2\text{NaCl+ 2H}_2\text{SO}_{4}}\xrightarrow{}\underset{\text{(Weaker acid)}}{2\text{NaHSO}_{4}+2\text{HCl}}$$

8.13. Identify the substance oxidised reduced, oxidising agent and reducing agent for each of the following reactions :

$$\textbf{(a)}\space2\textbf{AgBr(s)}+\textbf{C}_{6}\textbf{H}_{6}\textbf{O}_{2}\textbf{(aq)}\xrightarrow{}2\textbf{Ag(s)}+2\textbf{HBr(aq)}+\textbf{C}_{6}\textbf{H}_{4}\textbf{O}_{2}\textbf{(aq)}\\\textbf{(b)}\space\textbf{HCHO(1)}+2[\textbf{Ag(NH}_3\textbf{)}_2[^+\textbf{(aq)}+3\text{OH}^{\normalsize-}\textbf{(aq)}\xrightarrow{}2\textbf{Ag(s)}+\textbf{HCOO}^{\normalsize-}\textbf{(aq)}+4\textbf{NH}_3\textbf{(aq)}+2\textbf{H}_2\textbf{O (1)}\\\textbf{(c) HCHO (l) + 2Cu}^{2+}\textbf{(aq)}+5\textbf{OH}^{\normalsize-}\textbf{(aq)}\xrightarrow{}\textbf{Cu}_{2}\textbf{O(s)}+\textbf{HCOO}^{-}\textbf{(aq)}+3\textbf{H}_2\textbf{O (1)}\\\textbf{(d) N}_{2}\textbf{H}_{4}\textbf{(1)}+2\textbf{H}_{2}\textbf{O}_{2}\textbf{(1)}\xrightarrow{}\textbf{N}_{2}\textbf{(g)}+\textbf{4 H}_{2}\textbf{O (1)}\\\textbf{(e) \space Pb(s)}+\textbf{PbO}_{2}\textbf{(s)}+2\textbf{H}_{2}\textbf{SO}_{4}\textbf{(aq)}\xrightarrow{}2\textbf{PbSO}_{4}\textbf{(s)}+2\textbf{H}_2\textbf{O(1)}$$

Ans. (a) Oxidised substance → C6H6O2 (aq)

Reduced substance → AgBr(s)

Oxidising agent → AgBr(s)

Reducing agent → C6H6O2 (aq)

(b) Oxidised substance → HCHO (aq)

Reduced substance → [Ag(NH3)2]+ (aq)

Oxidising agent → [Ag(NH3)2]+ (aq)

Reducing agent → HCHO (aq)

(c) Oxidised substance → HCHO (aq)

Reduced substance → Cu2+(aq)

Oxidising agent → Cu2+(aq)

Reducing agent → HCHO (aq)'

(d) Oxidised substance → N2H4 (l)

Reduced substance → H2O2(l)

Oxidising agent → H2O2(l)

Reducing agent → N2H4(l)

(e) Oxidised substance → Pb(s)

Reduced substance → PbO2(s)

Oxidising agent → PbO2(s)

Reducing agent → Pb(s)

8.14. Consider the reactions:

2S2O32–(aq) + I2(s) → S4O62– (aq) + 2I(aq)

S2O32– (aq) + 2Br2(l) + 5H2O(l) → 2SO42–(aq) + 4Br(aq) + 10H+(aq)

What does the same reductant, thiosulphate react differently with iodine and bromine?

Ans. The average oxidation number of S in S2O32– is +2 while in S4O62– it is +2.5. The oxidation number of S in SO42– is +6. Since Br2 is a stronger oxidising agent than I2, it oxidises S of S2O32– to a higher oxidation state of +6 and hence forms SO42– ion. I2, however, being weaker oxidising agent oxidises S of S2O32– ion to a lower oxidation state of +2.5 in S4O62– ion. It is because of this reason that thiosulphate reacts differently with Br2 and I2.

8.15. Justify giving reactions that among halogens, fluorine is the best oxidant and amoing hydrohalic compounds, hydroiodic acid is the best reductant.

Ans. F2 can oxidise Cl to Cl2, Br to Br2, and I to I2 as:

$$\text{F}_{2}\text{(aq)}+2\text{Cl}^{\normalsize-}(s)\xrightarrow{}2\text{F}^{-}\text{(aq)}+\text{Cl(s)}\\\text{F}_{2}\text{(aq)}+2\text{Br}^{\normalsize-}\text{(aq)}\xrightarrow{}2\text{F}^{\normalsize-}\text{(aq)}+\text{Br}_{2}\text{(s)}\\\text{F}_{2}\text{(aq)}+2\text{I}^{\normalsize-}\text{(aq)}\xrightarrow{}2\text{F}^{\normalsize-}\text{(aq)}+\text{I}_{2}\text{(s)}$$

On the other hand, Cl2, Br2 and I2 cannot oxidise F to F2. The oxidising power of halogens increases in the order of I2 < Br2 < Cl2 < F2. Hence, fluorine is the best oxidant among halogens.

HI and HBr can reduce H2SO4 to SO2, but HCl and HF cannot. Therefore, HI and HBr are stronger reductants than HCl and HF.

$$2\text{HI} + \text{H}_2\text{SO}_4\xrightarrow{}\text{I}_{2}+\text{SO}_{2}+2\text{H}_{2}\text{O}\\2\text{HBr}+\text{H}_{2}\text{SO}_{4}\xrightarrow{}\text{Br}_{2}+\text{SO}_{2}+2\text{H}_{2}\text{O}$$

Again, I can reduce Cu2+ to Cu+, but Br cannot.

4I(aq) + 2Cu2+(aq) → Cu2I2(s) + I2(aq)

Hence, hydroiodic acid is the best reductant among hydrohalic compounds.

Thus, the reducing power of hydrohalic acids increases in the order of HF < HCl < HBr < HI.

8.16. Why does the following reaction occur?

XeO64–(aq) + 2F(aq) + 6H+(aq) → XeO3(g) + F2(g) + 3H2O(l)

What conclusion about the compound Na4XeO6 (of which XeO64– is a part) can be drawn from the reaction.

Ans. The given reaction occurs because XeO64– oxidises F and F reduces XeO64–.

$$\text{X}^{\normalsize+8}\text{eO}_{6}^{4-}(\text{aq})+2\text{F}^{\normalsize-1}\text{(aq)}+6\text{H}^{\normalsize+}(\text{aq})\xrightarrow{}\text{X}^{+6}\text{eO}_{3}\text{(g)}+\text{F}_{2}^{6}\text{(g)}+3\text{H}_{2}\text{O(1)}$$

In this reaction, the oxidation number (O.N.) of Xe decreases from +8 in XeO64– to +6 in XeO3 and the O.N. of F increases from –1 to F– to O in F2.

Therefore, XeO64– is reduced while F is oxidised.

Hence, we can conclude that the reaction occur because XeO64– a stronger oxidising agent than F2.

8.17. Consider the reactions :

(a) H3PO2(aq) + 4AgNO3(aq) + 2H2O(l) → H3PO4(aq) + 4Ag(s) + 4HNO3(aq)

(b) H3PO2(aq) + 2CuSO4(aq) + 2H2O(l) → H3PO4(aq) + 2Cu(s) + H2SO4(aq)

(c) C6H5CHO(l) + 2[Ag(NH3)2]+(aq) + 3OH–(aq) → C6H5COO–(aq) + 2Ag(s) + 4NH3(aq) + 2H2O(l)

(d) C6H5CHO(l) + 2Cu2+(aq) + 5OH(aq) → No change observed.

What inference do you draw about the behaviour of Ag+ and Cu2+ from these reactions?

Ans. Reactions (a) and (b) indicate that H3PO2 (hypophosphorous acid) is a reducing agent and thus reduces both AgNO3 and CuSO4 to Ag and Cu respectively. Conversely, both AgNO3 and CuSO4 act as oxidising agent and thus oxidise H3PO2 to H3PO4 (orthophosphoric acid).

Reaction (c) suggests that [Ag(NH3)2]+ oxidises C6H5CHO (benzaldehyde) to C6H5COO (benzoate ion)but reaction (d) indicates that Cu2+ ions cannot oxidise C6H5CHO to C6H5COO. Therefore, from the above reactions, we conclude that Ag+ ion is a strong oxidising agent than Cu2+ ion.

8.18. Balance the following redox reactions by ion – electron method :

(a) MnO4(aq) + I(aq) → MnO2(s) + I2(s) (in basic medium)

(b) MnO4(aq) + SO2(g) → Mn2+(aq) + HSO4–(aq) (in acidic solution)

(c) H2O2(aq) + Fe2+(aq) → Fe3+(aq) + H2O(l) (in acidic solution)

(d) Cr2O72– + SO2(g) → Cr3+ (aq) + SO42– (aq) (in acidic solution)

Ans. (a) Step 1 : The two half reactions involved in the given reaction are :

$$\text{Oxidation half reaction :}\text{I(aq)}^{\normalsize-1}\xrightarrow{}\text{I}_{2}^{0}\text{(s)}$$

Reduction half reaction :

$$\text{Mn}^{+7}\text{O}^{\normalsize-}_{4}\text{(aq)}\xrightarrow{}\text{Mn}^{\normalsize+4}\text{O}_{2}\text{(aq)}$$

Step 2 : Balancing I in the oxidation half reaction, we have :

$$2\text{I}^{-}\text{(aq)}\xrightarrow{}\text{I}_{2}\text{(s)}$$

Now, to balance the charge, we add 2e to the RHS of the reaction.

$$2\text{I}^{\normalsize–}\text{(aq)} \xrightarrow{} \text{I}_2\text{(s)} + 2e^{\normalsize–}$$

Step 3 : In the reduction half reaction, the oxidation state of Mn has reduced from +7 to +4.

Thus, 3 electrons are added to the LHS of the reaction.

$$\text{MnO}_{4}^{\normalsize-}\text{(aq)}+2\text{H}_{2}\text{O}+3e^{-}\xrightarrow{}\text{MnO}_{2}\text{(aq)}$$

Now, to balance the charge, we add 4 OH ions to the RHS of the reaction as the reaction is taking place in a basic medium.

$$\text{MnO}_{4}^{\normalsize-}\text{(aq)}+3e^{\normalsize-}\xrightarrow{}\text{MnO}_{2}\text{(aq)}+4\text{OH}^{\normalsize-}$$

Step 4 : In this equation, there are 6 O atoms on the RHS and 4 O atoms on the LHS. Therefore, two water molecules are added to the LHS.

$$\text{MnO}_{4}^{\normalsize-}\text{(aq)}+3\text{e}^{\normalsize-}\xrightarrow{}\text{MnO}_{2}\text{(aq)}+4\text{OH}^{-}$$

Step 5 : Equalising the number of electrons by multiplying the oxidation half reaction by 3 and the reduction half reaction by 2, we have :

$$6\text{I}^{\normalsize-}\text{(aq)}\xrightarrow{}3\text{I}_{2}\text{(s)}+6\text{e}^{\normalsize-}\\2\text{MnO}_{4}^{\normalsize-}\text{(aq)}+4\text{H}_{2}\text{O}+6\text{e}^{\normalsize-}\xrightarrow{}2\text{MnO}_{2}\text{(s)}+8\text{OH}^{-}\text{(aq)}$$

Step 6 : Adding the two half reactions, we have the net balanced redox reaction as :

$$6\text{I}^{\normalsize–}\text{(aq)} + 2\text{MnO}_4^{\normalsize–}(\text{aq}) + 4\text{H}_2\text{O(l)} \xrightarrow{} 3\text{I}_2\text{(s)} + 2\text{MnO}_2\text{(s)} + 8\text{OH}^{\normalsize–}(\text{aq})$$

(b) Following the steps as in part (a), we have the oxidation half reaction as :

$$\text{SO}_2\text{(g)} + 2\text{H}_2\text{O(l)} \xrightarrow{} \text{HSO}_{4}^{\normalsize–}(\text{aq}) + 3\text{H}^{+}(\text{aq}) + 2\text{e}^{\normalsize–}(\text{aq})\\\text{And the reduction half reaction as :}\\\text{MnO}_{4}^{-}\text{(aq)}+8\text{H}^{\normalsize+}\text{(aq)}+5\text{e}^{\normalsize-}\xrightarrow{}\text{Mn}^{2+}\text{(aq)}+4\text{H}_{2}\text{O}(1)$$

Multiplying the oxidation half reaction by 5 and the reduction half reaction by 2, and then by adding them, we have the net balanced redox reaction as :

$$2\text{MnO}_{4}^{–}(\text{aq}) + 5\text{SO}_2(\text{g}) + 2\text{H}_2\text{O(l)} + \text{H}^{\normalsize+}(\text{aq})→ 2\text{Mn}^{2+}(\text{aq}) + 5\text{HSO}_4^–\text{(aq)}$$

(c) Following the steps as in part (a), we have the oxidation half reaction as :

$$\text{Fe}^{2+}(\text{aq})\xrightarrow{}\text{Fe}^{3+}(\text{aq}) + e^{\normalsize–}\\\text{And the reduction half reaction as :}\\\text{H}_{2}\text{O(aq)}+2\text{H}^{\normalsize+}\text{(aq)}+2\text{(e)}^{\normalsize-}\xrightarrow{}2\text{H}_{2}\text{O(1)}$$

Multiplying the oxidation half reaction by 2 and then adding it to the reduction half reaction, we have the net balanced redox reaction as :

$$\text{H}_{2}\text{O}_{2}\text{(aq)}+2\text{Fe}^{2+}\text{(aq)}+2\text{H}^{+}\text{(aq)}\xrightarrow{}2\text{Fe}^{3+}\text{(aq)}+2\text{H}_{2}\text{O(1)}$$

(d) Following the steps as in part (a), we have the oxidation half reaction as :

$$\text{SO}_{2}\text{(g)}+2\text{H}_{2}\text{O(1)}\xrightarrow{}\text{SO}_{4}^{2-}\text{(aq)}+4\text{H}^{\normalsize+}\text{(aq)}+2\text{e}^{\normalsize-}\\\text{And the reduction half reaction as :}\\\text{Cr}_{2}\text{O}_{7}^{2-}\text{(aq)}+14\text{H}^{+}\text{(aq)}+6\text{e}^{-}\xrightarrow{}2\text{Cr}^{3+}\text{(aq)}+7\text{H}_{2}\text{O(1)}$$

Multiplying the oxidation half reaction by 3 and then adding it to the reduction half reaction, we have the net balanced redox reaction as :

$$\text{Cr}_2\text{O}_7^{2–}\text{(aq)} + 3\text{SO}_2\text{(g)} + 2\text{H}^+\text{(aq)}\xrightarrow{} 2\text{Cr}^{3+}(\text{aq}) + 3\text{SO}_4^{2–}(\text{aq}) + \text{H}_2\text{O(l)}$$

8.19. Balance the following equations in basic medium by ion-electron method and oxidation number methods and identify the oxidising agent and the reducing agent.

$$\textbf{(a) P}_4\textbf{(s) + OH}^{\normalsize–}\textbf{(aq)} \xrightarrow{} \textbf{PH}_3(\textbf{g) + HPO}_{2}^{\normalsize–}\textbf{(aq)}\\\textbf{(b) N}_2\textbf{H}_4\textbf{(l) + ClO}_3^{\normalsize–}(\textbf{aq})\xrightarrow{} \textbf{NO(g)} + \textbf{Cl}^{\normalsize–}\textbf{(g)}\\\textbf{(c) Cl}_2\textbf{O}_7\textbf{(g) + H}_2\textbf{O}_2\textbf{(aq)} \xrightarrow{} \textbf{ClO}_2^{\normalsize–} (\textbf{aq}) + \textbf{O}_2\textbf{(g) + H}^{\normalsize+}$$

Ans. (a)

The O.N. (oxidation number) of P decreases from 0 in P4 to –3 in PH3 and increases from 0 in P4 to +2 in HPO2–. Hence, P4 acts both as an oxidising agent and a reducing agent in this reaction.

Ion-electron method : The oxidation half electron is :

$$\text{P}_{4}\text{(s)}\xrightarrow{}4\text{HPO}_{2}^{\normalsize-}\text{(aq)}\\\text{This P atom is balanced as :}\\\text{P}_{4}^{0}\text{(s)}\xrightarrow{}4\text{H P}^{2+}\text{O}_{2}^{\normalsize-}\text{(aq)}\\\text{The O.N. is balanced by adding 8 electrons as :}\\\text{P}_{4}\text{(s)}\xrightarrow{}4\text{HPO}_{2}^{\normalsize-}\text{(aq)}+8e^{\normalsize-}\\\text{The charge is balanced by adding 12OH}^{\normalsize–} \text{as :}\\\text{P}_{4}\text{(s)}+12\text{OH}^{\normalsize-}\text{(aq)}\xrightarrow{}4\text{HPO}_{2}^{\normalsize-}\text{(aq)}+8\text{e}^{\normalsize-}\\\text{The H and O atoms are balanced by adding 4H}_2\text{O as:}\\\text{P}_{4}\text{(s)}+12\text{OH}^{\normalsize-}\text{(aq)}\xrightarrow{}4\text{HPO}_{2}^{\normalsize-}\text{(aq)}+\text{4H}_{2}\text{O(1)}+8\text{e}^{\normalsize-}\space...\text{(i)}$$

The reduction half equation is:

$$\text{P}_{4}\text{(s)}\xrightarrow{}\text{PH}_{3}\text{(g)}\\\text{The P atom is balanced as}\\\text{P}_{4}^{0}\text{(s)}\xrightarrow{}4\text{P}^{\normalsize-3}\text{H}_{3}\text{(aq)}\\\text{The O.N. is balanced by adding 12 electrons as:}\\\text{P}_{4}\text{(s)}+12e^{\normalsize-}\xrightarrow{}4\text{PH}_{3}\text{(g)}\\\text{The charge is balanced by adding 12OH}^{\normalsize–} \text{as :}\\\text{P}_{4}\text{(s)}+12\text{e}^{\normalsize-}\xrightarrow{}4\text{PH}_{3}\text{(g)}+12\text{OH(aq)}\\\text{The O and H atoms are balanced by adding 12H}_2\text{O as:}\\\text{P}_{4}\text{(s)}+12\text{H}_{2}\text{O}\text{(s)}+12e^{\normalsize-}\xrightarrow{}4\text{PH}_{3}\text{(g)}+12\text{HO}^{\normalsize-}\text{(aq)}$$

By multiplying equation (i) with 3 and (ii) with 2 and then adding them, the balanced chemical equation can be obtained as :

$$5\text{P}_4(\text{s}) + 12\text{H}_2\text{O(l)} + 12\text{HO}^{\normalsize–}(\text{aq}) \xrightarrow{} 8\text{PH}_3(\text{g}) + 12\text{HPO}_2^{\normalsize-} (\text{aq})$$

(b)

The oxidation number of N increases from –2 in N2H4 to +2 in NO and the oxidation number of Cl decreases from +5 in ClO3 to –1 in Cl. Hence, in this reaction, N2H4 is the reducing agent and ClO is the oxidising agent.

Oxidation number method :

Total decrease in oxidation number of
N = 2 × 4 = 8

Total increase in oxidation number of
Cl = 1 × 6 = 6

On multiplying N2H4 with 3 and ClO3 with 4 to balance the increase and decrease in oxidation number we get:

$$3\text{N}_2\text{H}_4\text{(l) + 4ClO}_3^{\normalsize–}(\text{aq}) \xrightarrow{} \text{NO(g) + Cl}^{–}\text{(aq)}\\\text{The N and Cl atoms are balances as:}\\ 3\text{N}_2\text{H}_4\text{(l)} + 4\text{ClO}_3^{–}\text{(aq)} → 6\text{NO(g) + 4Cl}^{\normalsize–}(aq) \\\text{The O atoms are balanced by adding 6H}_2\text{O as:}\\ 3\text{N}_2\text{H}_4\text{(l)} + 4\text{ClO}_3^–(\text{aq}) → 6\text{NO(g) + 4Cl}^{\normalsize–}\text{(aq) + 6H}_2\text{O(l)}$$

This is the required balanced equation.

Ion-electron method: The oxidation half equation is:

$$\text{N}_{2}^{\normalsize-2}\text{H}_{4}\text{(I)}\xrightarrow{}\text{N}^{+2}\text{O}\text{(g)}$$

The N atoms are balanced as :

$$\text{N}_2\text{H}_4\text{(l)} \xrightarrow{} 2\text{NO(g)}$$

The oxidation number is balanced by adding 8 electrons as:

$$\text{N}_2^{\normalsize-2}\text{H}_4\text{(l)}\xrightarrow{}\text{N}^{\normalsize+2}\text{O(g)}$$

The charge is balanced by adding 8 OH ions as:

$$\text{N}_{2}\text{H}_{4}(1)+8\text{OH}^{\normalsize-}\text{(aq)}\xrightarrow{}2\text{NO(g)}+8e^{\normalsize-}$$

The O atoms are balanced by adding 6H2O as :

$$\text{N}_{2}\text{H}_{4}(1)+8\text{OH}^{\normalsize-}\text{(aq)}\xrightarrow{}2\text{NO(g)}+6\text{H}_{2}\text{O(1)}+8e^{\normalsize-}\space\text{...(i)}\\\text{The reduction half equation is :}\\\text{Cl}^{\normalsize+5}\text{O}_{3}^{\normalsize-}\text{(aq)}\xrightarrow{}\text{Cl}^{^{\normalsize-}\normalsize-1}\text{(aq)}\\\text{The oxidation number is balanced by adding 6 electrons as :}\\\text{ClO}_3^{\normalsize–}(\text{aq}) + 6e^{\normalsize–}\xrightarrow{} \text{Cl}^{\normalsize–}(\text{aq})\\\text{The charge is balanced by adding 6OH}^{\normalsize–} \text{ions as :}\\\text{ClO}_{3}^{\normalsize-}\text{(aq)}+6e^{\normalsize-}\xrightarrow{}\text{Cl}^{\normalsize-}\text{(aq)}+6\text{OH}^{\normalsize-}\text{(aq)}\\\text{The O atoms are balanced by adding 3H}_2\text{O as:}\\\text{ClO}_{3}^{\normalsize-}\text{(aq)}+3\text{H}_{2}\text{O(1)}+6e^{\normalsize-}\xrightarrow{}\text{Cl}^{\normalsize-}\text{(aq)}+6\text{OH}^{\normalsize-}\text{(aq)}\space...\text{(ii)}\\\text{The balanced equation can be obtained by multiplying equation (i) with 3 and equation (ii) with 4 and then adding them as :}\\3\text{N}_{2}\text{H}_{4}(1)+4\text{ClO}_{3}^{\normalsize-}\text{(aq)}\xrightarrow{}6\text{NO(g)}+4\text{Cl}^{\normalsize-}\text{(aq)}+6\text{H}_{2}\text{O(1)}$$

The oxidation number of Cl decreases from +7 in Cl2O7 to +3 in ClO2 and the oxidation number of O increases from –1 in H2O2 to zero in O2. Hence, in this reaction, Cl2O7 is the oxidising agent and H2O2 is the reducing agent.

Oxidation number method :

Total decrease in oxidation number of Cl2O7 = 4 × 2 = 8

Total increase in oxidation number of H2O2 = 2 × 1 = 2

By multiplying H2O2 and O2 with 4 to balance the increase and decrease in the oxidation number, we get

8.20. What sorts of informations can you draw from the following reactions?

$$\textbf{(CN})_2(\textbf{g}) + 2\textbf{OH}^{\normalsize–}(\textbf{aq}) \xrightarrow{} \textbf{CN}^{\normalsize–}(\textbf{aq}) + \textbf{CNO}^{\normalsize–}(\textbf{aq}) + \textbf{H}_2\textbf{O(l)}$$

Ans. The reaction is a disproportionation reaction.

It occurs in basic medium.

The oxidation number of N in (CN)2, CN and CNO is –3, –2 and –5 respectively.

Cyanogen (CN)2 is simultaneously reduced to CN ion and oxidised to cyanate ion CNO ion.

8.21. The Mn3+ ion is unstable in solution and undergoes disproportionation to give Mn2+, MnO2 and H+ ion. Write a balanced ionic equation for the reaction.

Ans. The unbalanced chemical reaction is :

$$\text{Mn}^{3+}(\text{aq}) \xrightarrow{}\text{Mn}^{2+}(\text{aq}) + \text{MnO}_2\text{(s)} + \text{H}^{\normalsize+} (\text{aq})\\\text{The oxidation half reaction is,}\\\text{Mn}^{3+}\text{(aq)}\xrightarrow{}\text{MnO}_{2}\text{(s)}\\\text{To balance oxidation number, one electron is added on R.H.S.}\\\text{Mn}^{3+}\text{(aq)}\xrightarrow{}\text{MnO}_{2}\text{(s)}+\text{e}^{\normalsize-}\\\text{4 protons are added to balance the charge.}\\\text{Mn}^{3+}\text{(aq)}\xrightarrow{}\text{MnO}_{2}\text{(s)}+4\text{H}^{\normalsize+}\text{(aq)}+\text{e}^{\normalsize-}\\\text{2 water molecules are added to balance O atoms.}\\\text{Mn}^{3+}\text{(aq)}+2\text{H}_{2}\text{O}\xrightarrow{}\text{MnO}_{2}\text{(s)}+4\text{H}^{\normalsize+}\text{(aq)}+e^{\normalsize-}\space...\text{(i)}\\\text{The reduction half reaction is}\text{Mn}^{3+}\text{(aq)} \xrightarrow{} \text{Mn}^{2+}\text{(aq)}.\\\text{An electron is added to balance oxidation number.}\\\text{Mn}^{3+}\text{(aq)}+e^{\normalsize-}\xrightarrow{}\text{Mn}^{+2}\text{(aq)}\space...\text{(ii)}$$

Two half cell reactions are added to obtain balanced equation for disproportionation reaction.

$$2\text{Mn}^{3+}\text{(aq) + 2H}_2\text{O(l)} \xrightarrow{}\text{Mn}^{2+}\text{(aq) + MnO}_2\text{(s) + 4H}^{\normalsize+}\text{(aq)}$$

8.22. Consider the elements :

Cs, Ne, I and F

(a) Identify the element that exhibits only negative oxidation state.

(b) Identify the element that exhibits only positive oxidation state.

(c) Identify the element that exhibits both positive and negative oxidation states.

(d) Identify the element which exhibits neither the negative nor does the positive oxidation state.

Ans. (a) F exhibits only negative oxidation state of –1.

(b) Cs exhibits positive oxidation state of +1.

(c) I exhibits both positive and negative oxidation states. It exhibits oxidation states of –1, +1, +3, +5 and +7.

(d) The oxidation state of Ne is zero. It exhibits neither negative nor positive oxidation state.

8.23. Chlorine is used to purify drinking water. Excess of chlorine is harmful. The excess of chlorine is removed by treating with sulphur dioxide. Present a balanced equation for this redox change taking place in water.

Ans. The given redox reaction can be represented as:

$$\text{Cl}_2(\text{s) + SO}_2\text{(aq) + H}_2\text{O(l)}\xrightarrow{}\text{Cl(aq) + SO}_4^{2–}\text{(aq)}\\\text{The oxidation half reaction is :}\\\text{SO}_{2}^{\normalsize+4}\text{(aq)}\xrightarrow{}\text{S}^{\normalsize+6}\text{O}_{4}^{2-}\text{(aq)}\\\text{The oxidation number is balanced by adding two electrons as :}\\\text{SO}_{2}^{\normalsize+4}\text{(aq)}\xrightarrow{}\text{SO}_{4}^{2\normalsize-}\text{(aq)}+2e^{\normalsize-}\\\text{The charge is balanced by adding 4H}^{\normalsize+} \text{ions as :}\\\text{SO}_{2}\text{(aq)}+2\text{H}_{2}\text{O}_{4}^{2-}\text{(aq)}\xrightarrow{}4\text{H}^{\normalsize+}\text{(aq)}+2e^{\normalsize-}\\\text{The O atoms and H}^{+} \text{ions are balanced by adding 2H}_2\text{O molecules as :}\\\text{SO}_{2}\text{(aq)}+2\text{H}_{2}\text{O}_{4}^{2-}\text{(aq)}\xrightarrow{}4\text{H}^{\normalsize+}\text{(aq)}+2e^{\normalsize-}\space...(\text{i})\\\text{The reduction half reaction is :}\\\text{Cl}_{2}^{0}\text{(s)}\xrightarrow{}\text{Cl}^{\normalsize-^{\normalsize-1}}\text{(aq)}$$

The oxidation number is balanced by adding electrons

$$\text{Cl}_2\text{(s) + 2e}^{\normalsize–}\xrightarrow{}2\text{Cl}^{\normalsize-}\text{(aq)}\space...(\text{ii})\\\text{The balanced chemical equation can be obtained by adding equation (i) and (ii) as :}\\\text{Cl}_{2}(s)+\text{SO}_{2}\text{(aq)}+2\text{H}_{2}\text{O(1)}\xrightarrow{}2\text{Cl}^{\normalsize-}\text{(aq)}+\text{SO}_{4}^{2-}\text{(aq)}+4\text{H}^{\normalsize+}\text{(aq)}$$

8.24. Refer to the periodic table given in your book and now answer the following questions:

(a) Select the possible non metals that can show disproportionation reaction.

(b) Select three metals that can show dispropor-tionation reaction.

Ans. In disproportionation reactions, one of the reacting substances always contains an element that can exist in at least three oxidation states.

(a) P, Cl and S can show disproportionation reactions as these elements can exist in three or more oxidation states.

(b) Mn, Cu and Ga can slow disproportionation reactions as these element can exist in three or more oxidation states.

8.25. In Ostwald’s process for the manufacture of nitric acid, the first step involves the oxidation of ammonia gas by oxygen gas to give nitric oxide gas and steam.

What is the maximum weight of nitric oxide that can be obtained starting only with 10.00 g. of ammonia and 20.00 g of oxygen?

Ans. The balanced chemical equation for the given reaction is given as :

$$\underset{=68g}{\underset{4×17g}{4\text{NH}_{3}\text{(g)}}}+\underset{=160g}{\underset{5×32g}{5\text{O}_2\text{(g)}}}\xrightarrow{}\underset{=120g}{{\underset{4×30g}{4\text{NO(g)}}}}+\underset{=108g}{\underset{4×30g}{6\text{H}_{2}\text{O(g)}}}$$

Thus, 68 g of NH3 reacts with 160 g of O2.

$$\text{Therefore, 10g of NH}_3 \text{reacts with}\frac{160×10}{68}\text{g}\space\text{of}\space\text{O}_{2},\text{or 25.53 g of O}_2.$$

or 25.53 g of O2.

But the available amount of O2 is 20 g.

Therefore, O2 is the limiting reagent (we have considered the amount of O2 to calculate the weight of nitric oxide obained in the reaction).

Now, 160 g of O2 gives 120 g of NO.

$$\text{Therefore, 20 g of O}_2 \text{gives}\space\frac{120×20}{160}\space\text{g of N, or 15 g of NO.}$$

Hence, a maximum of 15 g of nitric oxide can be obtained.

8.26. Using the standard electrode potentials given in the Table 8.1, predict if the reaction between the following is feasible:

(a) Fe3+(aq) and I(aq)

(b) Ag+(aq) and Cu(s)

(c) Fe3+ (aq) and Cu(s)

(d) Ag(s) and Fe3+(aq)

(e) Br2(aq) and Fe2+(aq)

Ans. (a) When emf is positive, the reaction is feasible.

cell = E°Fe-E°I2

= 0.77 – 0.54 = 0.23 V

Reaction is feasible.

(b) E°cell = E°Ag-E°Cu

= 0.80 – 0.34 = 0.46 V

Reaction is feasible.

(c) E°cell = E°Fe-E°Cu

= 0.77 – 0.34 = 0.33 V

Reaction is feasible.

(d) E°cell = E°Fe-E°cu

= 0.77 – 0.80 = –0.03 V

Reaction is not feasible.

(e) E°cell = E°Br2-E°Fe

= 1.09 – 0.77 = 0.32 V

Reaction is feasible.

8.27. Predict the products of electrolysis in each of the following:

(i) An aqueous solution of AgNO3 with silver electrodes

(ii) An aqueous solution AgNO3 with platinum electrodes

(iii) A dilute solution of H2SO4 with platinum electrodes

(iv) An aqueous solution of CuCl2 with platinum electrodes.

Ans. (i) AgNO3 ionizes in aqueous solutions to form Ag+ and NO3ions.

On electrolysis, either Ag+ ions of H2O molecules can be reduced at the cathode. But the reduction potential of Ag+ ions is higher than that of H2O.

$$\text{Ag}^{\normalsize–}(\text{aq}) + \text{e}^{\normalsize–} \xrightarrow{} \text{Ag(s)}; \text{E}°=+0.80\text{V}\\2\text{H}_2\text{O(l)} + 2\text{e}^{–} \xrightarrow{} \text{H}_2\text{(g)} + 2\text{OH}^{\normalsize–}(\text{aq});\text{E}° = –0.83\text{V}$$

Hence, Ag+ ions are reduced at the cathode. Similarly, either Ag metal of the anode or H2O molecular may be oxidised at the anode. But the oxidation potential of Ag is higher than that of H2O molecules.

$$\text{Ag(s)}\xrightarrow{}\text{Ag}^{\normalsize+}\text{(aq)}+e^{\normalsize-};\space\text{E°}=-0.80\text{V}\\2\text{H}_{2}\text{O(1)}\xrightarrow{}\text{O}_{2}\text{(g)}+4\text{H}^{+}\text{(aq)}+4e^{\normalsize-}; \text{E}°=-1.23\text{V}$$

Therefore, Ag metal gets oxidised at the anode. It may, however, be mentioned here that the oxidation potential of NO3 ions is even lower than that of H2O since more bonds are to be broken during reduction of nitrate ions than those in H2O. Thus, when an aqueous solution of AgNO3 is electrolysed, silver metal from Ag anode dissolves while Ag+(aq) ions present in the solution get reduced and get deposited on the cathode.

(ii) If electrolysis of AgNO3 solution is carried out using platinum electrodes, istead of silver electrodes, oxidation of water occurs at anode to liberate oxygen. This is because Pt being a noble metal does not undergo oxidation easily. Whereas, Ag+ ions are reduced and get deposited at the cathode. Therefore, when an aqueous solution of AgNO3 is electrolysed using platinum electrodes, Ag+ ions from the solution get deposited on the cathode while O2 is liberated as the anode.

(iii) H2SO4 ionizes in aqueous solutions to give H+ and SO42– ions.

$$\text{H}_{2}\text{SO}_{4}\text{(aq)}\xrightarrow{}2\text{H}^{+}\text{(aq)}+\text{SO}_{4}^{2-}\text{(aq)}$$

On electrolysis, either of H+ ions or H2O molecules can get reduced at the cathode. But the reduction potential of H+ ions is higher than that of H2O molecules.

$$2\text{H}^{+}\text{(aq)} + 2e– \xrightarrow{} \text{H}_2(\text{g});\space\text{E}° = 0.0\text{V}\\2\text{H}_2\text{O(aq)} + 2e^{\normalsize–} \xrightarrow{} \text{H}_2(\text{g}) + 2\text{OH}^{\normalsize–}(\text{aq});\text{E}° = –0.83\text{V}$$

Hence, at the cathode, H+ ions are reduced to liberate H2 gas.

On the other hand at the anode, either of SO42– ions or H2O molecules can get oxidised.
But the oxidation of SO42– involves breaking of more bonds than that of H2O molecules.

Hence, SO42– ions have a lower oxidation potential than H2O. Thus, H2O is oxidised at the anode to liberate O2 molecule.
Thus, during electrolysis of aqueous solution of H2SO4, only the electrolysis of H2O occurs liberating H2 at the cathode and O2 at the anode.

(iv) In aqueous solutions, CuCl2 ionises to give Cu2+ and Cl ions as :

$$\text{CuCl}_{2}\text{(aq)}\xrightarrow{}\text{Cu}^{2-}\text{(aq)}+2\text{Cl}(\text{aq})$$

On electrolysis, either of Cu2+ ions or H2O molecules can get reduced at the cathode. But the reduction potential of Cu2+ is more than that of H2O molecules.

$$\text{Cu}^{2+}(\text{aq}) + 2e– \xrightarrow{}\text{Cu(aq)} ; \text{E}° = +0.34\text{V}\\ \text{H}_2\text{O(l)} + 2e^{\normalsize–} \xrightarrow{} \text{H}_2(\text{g}) + 2\text{OH}^{\normalsize–}; \text{E}° = –0.83 \text{V}$$

Hence, Cu2+ ions are reduced at the cathode and get deposited.

Similarly, at the anode, either of Cl or H2O is oxidised. The oxidation potential of H2O is higher than that of Cl.

$$2\text{Cl}^{–}\text{(aq)} \xrightarrow{} \text{Cl}_2(\text{g}) + 2e– ; \text{E}° = –1.36\text{V}\\2\text{H}_2\text{O(l)} \xrightarrow{} \text{O}_2(\text{g}) + 4\text{H}^+(\text{aq}) + 4e–; \text{E}° = –1.23 \text{V}$$

But oxidation of H2O molecules occurs at a lower electrode potential than that of Cl ions because of over-voltage (extra voltage required to liberate gas). As a result, Cl ions are oxidized at the anode to liberate Cl2 gas.

Thus, when an aqueous solution of CuCl2 is electrolysed, Cu metal is deposited at the cathode while Cl2 gas is evolved at the anode.

8.28. Arrange the following metals in the order in which they displace each other from the solution of their salts.

Al, Cu, Fe, Mg and Zn.

Ans. A metal of stronger reducing power displaces another metal of weaker reducing power from its solution of salt.

The order of the increasing reducing power of the given metals is Cu < Fe < Zn < Al < Mg.

Hence, we can saythat Mg can displace Al from its salt solution, but Al cannot displace Mg.

Thus, the order in which the given metals displace each other from the solution of their salts is given below :

Mg > Al > Zn > Fe > Cu

8.29. Given the standard electrode potentials,

K+/K = –2.93V, Ag+/Ag = 0.80V.

Hg2+/Hg = 0.79V

Mg2+/Mg = –2.37V.Cr3+/Cr = –0.74 V

arrange these metals in their increasing order of reducing power.

Ans. The lower the electrode potential, the stronger is the reducing agent. Therefore, the increasing order of the reducing power of the given metals is :

Ag < Hg < Cr < Mg < K

0.80V < 0.79V < –0.74V < – 2.37V < – 2.93 V

8.30. Depict the galvanic cell in which the reaction

Zn(s) + 2Ag+(aq) → Zn2+(aq) + 2Ag(s) takes place, Further show:

(i) which of the electrode is negatively charged.

(ii) the carriers of the current in the cell, and

(iii) individual reaction at each electrode.

Ans. In the given redox reaction,

$$\text{Zn(s) + 2Ag}^{+}(\text{aq})\xrightarrow{} \text{Zn}^{2+}(\text{aq}) + 2\text{Ag(s)}$$

Oxidation occurs at zinc electrode because Zn is getting oxidised to Zn2+ ions and reduction occurs at the silver electrode because Ag+ ions are reduced to Ag metal. Therefore,

The galvanic cell corresponding to the given redox reaction can be represented as :

Zn|Zn+2(aq) || Ag+(aq) |Ag

(i) Zn electrode is negatively charged because at this electrode, Zn oxidises to Zn2+ and the leaving electrons accumulate on this electrode.

(ii) Ions are the carriers of at current in the cell.

(iii)  The reaction taking place at Zn electrode can be represented as :

$$\text{Zn(s)} \xrightarrow{} Zn^{2+}(\text{aq}) + 2e^{\normalsize–}\\\text{And the reaction taking place at Ag electrode can be represented as :}\\\text{Ag}^{+}\text{(aq)}+e^{-}\xrightarrow{}\text{Ag(s)}$$

Similarly, at the anode, either of Cl or H2O is oxidised. The oxidation potential of H2O is higher than that of Cl.

$$2\text{Cl}^{-}\text{(aq)}\xrightarrow{}\text{Cl}_{2}\text{(g)}+2e^{\normalsize-};\space\text{E°}=-1.36\text{V}\\2\text{H}_{2}\text{O}\xrightarrow{}\text{O}_{2}\text{(g)}+4\text{H}^{\normalsize+}\text{(aq)}+4e^{\normalsize-};\text{E°}=-1.23\text{V}$$

But oxidation of H2O molecules occurs at a lower electrode potential than that of Cl ions because of over-voltage (extra voltage required to liberate gas). As a result, Cl ions are oxidised at the anode to liberate Cl2 gas.